Question

ich has the highest energy? 7. a) Depending upon the kind of models you are using, you may or may not be able to construct models of cyclopropane and cyclobutane. Do so, if possible, but do not break the linkages. Why do we say that these molecules are strained? What is the normal sp3 bond angle? What are the (internal) C-C-C bond angles in cyclopropane? b) Make a model of cyclopentane. Does this molecule seem to have any ring strain in the sense that the three and four-membered rings do? The molecule is not completely strain-free because there is some eclipsing. Identify the eclip sed atoms or groups when the molecule is planar. In fact the molecule is not planar, but is puckered in either of two ways: i) One C-atom is above or below the plane defined by the other four envelope conformation. an coplanar C-atoms defining

Answer 1

Solution We know that C-atom makes 4 bonds and in ideal case the bond ample is 109.5° In case of cyclopropane and cyclobutane, the C-C-C bond angle is 60° and 90° respectively. cyclobutane сугорорта These bond angles can be calculated wing formula 180(n-2), where nis No.of C-atomo forming the ring For Ef: In case of cyclopropane n=3 180(n-2) = 180(3-2) = 180 = 60°. In 3 3 Now coming to the cyclobutane, the H-atoms are at eclipsing positions when the cyclobutane ring is planar. 0 I propeller shape - planar (reduction in strain). Yo reduce this strain due to eclipsing of H-atoms., cyclobutane I adoph a shape called propeller shape, in which there is reduction in the eclipsing between H-atoms. In this shape the two diagonal C-atoms move up the plane and other two (-atorns move down the plane.

Coming to your next question about cyclopentane, fint thing willich is Jo be noted is the C-C-C Angeld in polopenderro, is 108° which is very close to the tetrahedral angle of 109.5%. So, there is nol no angle strain , but there is a strain arising due to the eclipsing of groups attached to the l-alons. This type of straim is called forsional straim. Spo, cohen the indecule is planar, there exist torsional strain due to the eclipsing of H-atoms in eycloperutame ring. 20 H-alony Due to this eclipsing, cyclopentane odloph to two conformations, first envelope conformation in which, one of the C-atom moves above or below the plane elsfired by other 4 Boban C-atoms. envelope / No eclipsing) conformation of t-alons !). 1. The other conformation is called half-chair conformation in which one C-atom moves above the plane and other an adjacent C-atom moves below the plane defined by the remaining 3 C-atoms.

Half-Chair conformation (Reduction in ecliping) Due to these two conformations, there is reduction in the torsional strain due to reduction in the eclipsing of atonus or groups. Also, these two conformations rcupiclly interconwert into each other. envelope half-Chair. Now for your final question, ie, structure of cis- and bremer come a tran's-1,2-dimethyleyclopentane 14 V the me H trans. cis In case of cis-form there exists a plane passing through the middle of the compound, making dividing it into hoo oqual halves. This plane is also called mirror plane and so eis-form is a meso form.

i minor plane cis-form (meso). {achiral] 1. But in case of trans-form there is does not exist Such mirror plane'. Also, trans- form is chival because there is no symmetry aniinoy - - - - - 1 -- - и ме ме и. Non-superimposable mirror images. trans - form . [chival].